In the figure,if angle ACB = angle CDA,AC = 8 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $AC = 8 \, cm$,$AD = 3 \, cm$ and $\angle ACB = \angle CDA = 90^{\circ}$ (from the figure).In right-angled $	riangle ADC$,by Pythagoras th

Vedclass · Accepted Answer

$\frac{55}{3}$

Vedclass · Answer

(A) Given: $AC = 8 \, cm$,$AD = 3 \, cm$ and $\angle ACB = \angle CDA = 90^{\circ}$ (from the figure).In right-angled $	riangle ADC$,by Pythagoras theorem:$AC^2 = AD^2 + CD^2$$8^2 = 3^2 + CD^2$$64 = 9 + CD^2$$CD^2 = 55$$CD = \sqrt{55} \, cm$.Now,consider $	riangle CDB$ and $	riangle ADC$:$1$. $\angle BDC = \angle ADC = 90^{\circ}$ (given).$2$. $\angle DBC = \angle DCA$ (since both are equal to $90^{\circ} - \angle A$).By $AA$ similarity criterion,$	riangle CDB \sim 	riangle ADC$.Therefore,the ratio of corresponding sides is equal:$\frac{CD}{BD} = \frac{AD}{CD}$$CD^2 = AD 	imes BD$$55 = 3 	imes BD$$BD = \frac{55}{3} \, cm$.

In the figure,if $\angle ACB = \angle CDA$,$AC = 8 \, cm$ and $AD = 3 \, cm$,find $BD$ (in $cm$).

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

$A$ person begins his journey from point $A$. From $A$,he moves $800 \, m$ towards north to reach point $B$. From $B$,he moves $500 \, m$ towards east to reach point $C$. From $C$,he moves $400 \, m$ towards north to reach point $D$. Find the direct distance in $m$ from $A$ to $D$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $a = 9$ and $c = 12$,then $b = \dots$

Prove that the sum of the squares of the sides of a rectangle is equal to the sum of the squares of its diagonals.

Vedclass Test Series

Exam Paper Generator

Online Exam Module